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MARKS/PUNTE: 150
These marking guidelines consist of 19 pages./ Hierdie nasienriglyne bestaan uit 19 bladsye.
PHYSICAL SCIENCES: PHYSICS (P1) FISIESE WETENSKAPPE: FISIKA (V1)
NOVEMBER 2020
MARKING GUIDELINES/NASIENRIGLYNE
GRADE/GRAAD 12
SENIOR CERTIFICATE/SENIOR SERTIFIKAAT NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT
Physical Sciences/P1/Fisiese Wetenskappe/V1 2 DBE/November 2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
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QUESTION 1/VRAAG 1 1.1 B (2) 1.2 D (2) 1.3 C (2) 1.4 C (2) 1.5 C (2) 1.6 A (2) 1.7 A (2) 1.8 D (2) 1.9 A (2) 1.10 B (2)
[20]
Physical Sciences/P1/Fisiese Wetenskappe/V1 3 DBE/2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
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w
N
f T
Fapplied
QUESTION 2/VRAAG 2
2.1 Marking criteria/Nasienriglyne If any of the underlined key words/phrases in the correct context are omitted: - 1 mark per word/phrase. Indien enige van die sleutelwoorde/frases in die korrekte konteks weggelaat word: - 1 punt per woord/frase
The perpendicular force exerted by a surface on an object in contact with the surface. Die loodregte krag deur 'n oppervlak uitgeoefen op 'n voorwerp wat daarmee in kontak is. (2)
2.2
Accepted symbols/Aanvaarde simbole
N FN/Normal/Normal force/173,5N /Normaal/Normaalkrag
f Ff /fk /frictional force/wrywingskag/kinetic frictional force/kinetiese wrywingskrag/5 N
w Fg /mg/Weight/FEarth on block/Fw/Gewig/Gravitational force/ Gravitasiekrag/196 N
T Tension/Spanning/FT
Fapplied Ftoegepas
F/Applied force/35 N/Toegepaste krag/ FA
Notes/Aantekeninge
Mark is awarded for label and arrow./Punt word toegeken vir byskrif en pyltjie.
Do not penalise for length of arrows./Moenie vir die lengte van die pyltjies penaliseer nie.
Deduct 1 mark for any additional force./Trek 1 punt af vir enige addisionele krag. If all forces are correctly drawn and labelled, but no arrows, deduct 1 mark. /
Indien all kragte korrek geteken en benoem is, maar geen lyne nie, trek 1punt af.
(5)
2.3 For the/Vir die 20 kg: Fnet = ma T – f – FAx = ma T – 5 – 35 cos40° = 0 T = 31,81 N For/vir m: Fnet = ma mg – T = ma m(9,8) – 31,81 = 0 m = 3,25 kg
Marking criteria/Nasienriglyne
Formula for 20 kg or m kg/Formule vir 20 kg of m kg / Fnet = ma
Substitution of zero into either formula Vervanging van nul in een van die formules
All substitutions into Fnet for 20 kg as shown Alle vervanging in Fnet for 20 kg soos getoon
Substitution of value of T in eqn for m /Substitusie van waarde vir T in vgl vir m
Final answer/finale antwoord: 3,25 kg
(5)
2.4.1 Decreases/Neem af (1)
Physical Sciences/P1/Fisiese Wetenskappe/V1 4 DBE/2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
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2.4.2 POSITIVE MARKING FROM QUESTION 2.3
POSITIEWE NASIEN VANAF VRAAG 2.3
Moving to the right/Beweeg na regs
Velocity decreases/snelheid neem af Accelerates/Net force to left /Versnelling/netto krag na links OR/OF As the tension force decreases, the net force/acceleration acts in the opposite direction of motion /to the left. Soos die spanning afneem, is daar ‘n netto krag/versnelling in die teenoorgestelde rigting / na links Moving to the left/Beweeg na links Velocity increases/snelheid neem toe Accelerates/Net force to left /Versnelling/netto krag na links
(3) [16] QUESTION 3/VRAAG 3 3.1 (Motion of an object) under the influence of gravity (weight) only. (2 or 0)
(Beweging van ʼn voorwerp) slegs onder die invloed van gravitasie (gewig). OR/OF (Motion in which) the only force acting on the object is gravity (weight). (Beweging waar) die enigste krag wat op die voorwerp inwerk, gravitasie (gewig) is.
(2) 3.2.1 Δt = 0,67 – 0,64 = 0,03 s (2)
3.2.2 OPTION 1/OPSIE 1
Δt = 2
0,67)(1,90
= 0,62 s (0,615 s)
OPTION 2/OPSIE 2 Δx = viΔt + ½ aΔt2 (-1,85) = 0 + ½ (-9,8)Δt2 Δt = 0,61 s (0,6145 s)
(2)
OPTION 3/OPSIE 3
Δt = 2
0,67)(1,90= 1,285 s
Δt = 1,285 – 0,67 = 0,62 s (0,615 s)
OPTION 4/OPSIE 4 vf
2 = vi2 + 2aΔx
0 = vi2 + 2(-9,8)(1,85)
vi = 6,02 m·s-1 vf = vi
+ aΔt 0 = 6,02 + (-9,8)Δt Δt = 0,61 s
Physical Sciences/P1/Fisiese Wetenskappe/V1 5 DBE/2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
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3.2.3 POSITIVE MARKING FROM QUESTION 3.2.2 POSITIEWE NASIEN VANAF VRAAG 3.2.2
Marking Criteria/Nasienriglyne
Any appropriate formula/Enige geskikte formule
Correct substitution/Korrekte vervanging
Final answer/Finale antwoord: 5,94 to 6,08 m·s-1
OPTION 1/OPSIE 1 Upwards positive/Opwaarts positief vf = vi + aΔt 0 = vi + (-9,8)(0,62) vi = 6,08 m·s-1
(6,076 m·s-1 )
OPTION 2/OPSIE 2 Upwards positive/Opwaarts positief Δy = viΔt + ½ aΔt2 1,85 = vi (0,62) + ½ (-9,8) (0,62)2
vi = 6,02 m·s-1
(6,022 m·s-1 )
Downwards positive/Afwaarts positief vf = vi + aΔt 0 = vi + (9,8)(0,62) vi = -6,08 6,08 m·s-1
(6,076 m·s-1 )
Downwards positive/Afwaarts positief Δy = viΔt + ½ aΔt2
1,85 = vi (0,62) + ½ (9,8) (0,62)2
vi = -6,02 vi = 6,02 m·s-1 (6,022 m·s-1)
OPTION 3/OPSIE 3 Motion from top to bottom / Beweging vanaf bo na onder Downwards positive/Afwaarts positief
ya2vv2
i
2
f
vf2 = 0 + 2(9,8)(1,85)
vf = 6,02 m·s-1
initial velocity/beginsnelheid=6,02 m·s-1 Upwards positive/Opwaarts positief
ya2vv2
i
2
f
vf2 = 0 + 2(-9,8)(-1,85)
vf = 6,02 m·s-1
initial velocity/beginsnelheid=6,02 m·s-1 Motion from bottom to top Beweging vanaf onder na bo Downwards positive/Afwaarts positief
ya2vv2
i
2
f
02 = vi2 + 2(9,8)(-1,85)
vi = 6,02 m·s-1
Upwards positive/Opwaarts positief
ya2vv2
i
2
f
0 = vi2 + 2(-9,8)(1,85)
vi = 6,02 m·s-1
OPTION 4/OPSIE 4 Upwards positive/Opwaarts positief Δy = viΔt + ½ aΔt2 0 = vi(1,23) + ½ (-9,8)(1,23)2 vi = 6,03 m·s-1 Downwards positive/Afwaarts positief Δy = viΔt + ½ aΔt2 0 = vi(1,23) + ½ (9,8)(1,23)2 vi = - 6,03 m·s-1 speed/spoed = 6,03 m·s-1
OPTION 5/OPSIE 5
Δy =
2
vv if Δt
1,85 =
2
v0 i (0,62)
vi = 5,97 m·s-1
OPTION 6/OPSIE 6 FnetΔt = mΔv FnetΔt = m(vf – vi) m(9,8)(0,62) = m(0 – vi) vi = 6,08 m·s-1
OPTION 7/OPSIE 7 (Ep + Ek)floor/vloer = (Ep + Ek)top/bo (mgh + ½ mv2)floor/vloer = (mgh + ½ mv2)top/bo 0 + ½ v2 = (9,8)(1,85) + 0 v = 6,02 m·s-1
(3)
Physical Sciences/P1/Fisiese Wetenskappe/V1 6 DBE/2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
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3.2.4 OPTION/OPSIE 1, 2, 3, 4: Marking criteria/Nasienriglyne Calculate initial velocity:
Bereken aanvanklike snelheid:
Appropriate formula/Geskikte formule
Substitution/Vervanging
Calculate/Bereken Δt:
Appropriate formula/Geskikte formule
Substitution/Vervanging
1,97 s + Δt
Fin answer/Fin antwoord: 2,95 – 2,97 s
Calculate initial velocity: Bereken beginsnelheid
Calculate time Δt Bereken tyd Δt
OPTION 1/OPSIE 1 Downwards positive/Afwaarts positief
ya2vv2
i
2
f
0 = vi2 + 2(9,8)(-1,2)
vi = - 4,85 m·s-1
Upwards positive/Opwaarts positief
ya2vv2
i
2
f
0 = vi2 + 2(-9,8)(1,2)
vi = 4,85 m·s-1
Upwards positive Opwaarts positief Δy = viΔt + ½ aΔt2 1,2 = (4,85)Δt + ½(-9,8)Δt 2
Δt = 0,4898 s / 0,5 s t = 1,97 + 2(0,4898) = 2,95 s / 2,97 s OR/OF Δy = viΔt + ½ aΔt2
0 = (4,85)Δt + ½(-9,8)Δt2
Δt = 0,9898 s (or Δt = 0) t = 1,97 + 0,9898 = 2,96 s
Downwards positive Afwaarts positief Δy = viΔt + ½aΔt2 1,2 = (-4,85)Δt + ½(9,8)Δt2
Δt = 0,4898 s / 0,5 s t = 1,97 + 2(0,4898) = 2,95 s / 2,97 s OR/OF Δy = viΔt + ½aΔt2
0 = (4,85)Δt + ½(9,8)Δt 2 Δt = 0,9898 s (or Δt = 0) t = 1,97 + 0,9898 = 2,96 s OR/OF vf = vi + aΔt
-4,85 = 4,85 + (-9,8)Δt Δt = 0,9898 s Δt = 1,97 + 0,9898 = 2,96 s OR/OF Upwards positive Opwaarts positief vf = vi + aΔt
0 = 4,85 + (-9,8)Δt Δt = 0,4949 s Δt = 1,97 + (2)(0,4949) = 2,96 s OR/OF
Δy =
2
vv fi Δt
1,2 =
2
85,40ΔtΔt = 0,4948s
Δttotal = 2(0,4948) = 0,99 s Δt = 1,97 + 0,99 = 2,96 s
OPTION 2/OPSIE 2 (Emech)top = (Emech)bot/ond (Ep +Ek)top = (Ep +Ek)Bot/Ond
(mgh + ½mv2)top = (mgh + ½mv2)Bot/Ond (9,8)(1,2) + 0 = 0 + (½)v2 vi
= 4,85 m·s-1 upwards /opwaarts
OPTION 3/OPSIE 3 Wnc = ΔEp + ΔEk
0 = (0 – mgh) + ½m( 2
i
2
f vv )
0 = -(9,8)(1,2) + ½vi2
vi = 4,85 m·s-1 upwards /opwaarts
OPTION 4/OPSIE 4 Wnet = ΔEk
wΔxcos180° = ½m( 2
i
2
f vv )
(9,8)(1,2)cos180° = ½vi2
vi = - 4,85 m·s-1
Any one/ Enige een
Any one/ Enige een
Any one/ Enige een
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OPTION 5/OPSIE 5 Downwards positive/Afwaarts positief Δy = viΔt + ½ aΔt2
1,2 = 0 + ½(9,8) Δt2 Δt = 0,49 s t = 1,97 + 2(0,49) = 2,96 s Upwards positive/Opwaarts positief Δy = viΔt + ½ aΔt2
-1,2 = 0 + ½(-9,8)Δt2 Δt = 0,49 s t = 1,97 + 2(0,49) = 2,96 s
OPTION 5: Marking criteria/ OPSIE 5: Nasienriglyne
Formula /Formule
Substitution/Vervanging Δy = 1,2
Substitution/Vervanging 0 + ½(9,8) Δt2
1,97 s +
2 Δt
Final answer/Finale antwoord: 2,95 - 2,97 s
(6)
[15] QUESTION 4/VRAAG 4 4.1 (Linear) momentum (of an object) is the product of mass and velocity.
(Liniêre) momentum (van 'n voorwerp) is die produk van massa en snelheid. (2 or/of 0)
(2)
4.2.1 OPTION 1/OPSIE 1 East as positive/Oos as positief ∑pi = ∑pf mpvpi + mQvQi = mpvpf + mQvQf (0,16)(10) + (0,2)(-15) = (0,16)(-5) +(0,2)vQf vQf = -3 m∙s-1
vQf = 3 m∙s-1 west/wes
OPTION 2/OPSIE 2 West as positive/Wes as positief ∑pi = ∑pf mpvpi + mQvQi = mpvpf + mQvQf (0,16)(-10) + (0,2)(15) = (0,16)(5) +(0,2)QNf vQf = 3 m∙s-1
west/wes
OPTION 3/OPSIE 3 Δpp = -ΔpQ (0,16)(-5 – 10) = -(0,2)(v – (-15)) v = -3 m·s-1 = 3 m·s-1 west/wes
(5)
Any one/Enige een
Any one/Enige een
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4.2.2 For ball/Vir bal P: West as negative/Wes as negatief Impulse = Δp FnetΔt = Δp
Δp = m(vPf – vPi) = 0,16(-5 – 10) = - 2,4 2,4 N∙s (2,4 kg∙m∙s-1) OR/OF West as positive /Wes as positief Impulse = Δp FnetΔt = Δp
= m(vPf – vPi) = 0,16(5 – (-10)) = 2,4 N·s
POSITIVE MARKING FROM QUESTION 4.2.1 /POSITIEWE NASIEN VANAF VRAAG 4.2.1
(3)
For ball/Vir bal Q: West as negative/Wes as negatief Impulse = Δp FnetΔt = Δp
= m(vQf – vQi) = 0,2[-3 – (-15)] = 2,4 N∙s (2,4 kg∙m∙s-1) OR/OF West as positive /Wes as positief Impulse = Δp FnetΔt = Δp
= m(vQf – vQi) = 0,16(3 – (15)) = - 2,4 N·s 2,4 N∙s (2,4 kg∙m∙s-1)
[10] QUESTION 5/VRAAG 5
5.1 Marking criteria/Nasienriglyne If any of the underlined key words/phrases in the correct context are omitted: - 1 mark per word/phrase. However, IF: The word “work” is omitted 0 marks Indien enige van die sleutelwoorde/frases in die korrekte konteks weggelaat word: - 1 punt per woord/frase. Maar, INDIEN: Die woord “arbeid” uitgelaat is, 0 punte
A force is non-conservative if the work it does on an object (which is moving between two points) depends on the path taken. 'n Krag is nie-konserwatief indien die arbeid wat dit verrig (op 'n voorwerp wat tussen twee punte beweeg) afhanklik is van die pad.
OR/OF A force is non-conservative if the work it does on an object depends on the path taken. 'n Krag is nie-konserwatief indien die arbeid wat dit verrig afhanklik is van die pad.
OR/OF A force is non-conservative if the work it does in moving an object around a closed path is non-zero. 'n Krag is nie-konserwatief indien die arbeid wat dit verrig om 'n voorwerp op 'n geslote pad te beweeg, nie-nul is nie.
(2)
5.2
K = ½ mv2 /Ek = ½ mv2 ΔK = Kf - Ki
K = ½mvf2 - ½mvi
2 = ½m(vf
2- vi2)
=½(200)(22 - 42)
K = - 1 200 J
(3)
Any one/ Enige een
Any one/ Enige een
Any one/ Enige een
Any one/ Enige een
Any one /Enige een
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5.3 POSITIVE MARKING FROM QUESTION 5.2. POSITIEWE NASIEN VANAF VRAAG 5.2.
Marking criteria/Nasienriglyne
Appropriate formula/Geskikte formule
Substitution into appropriate formula together with/Vervanging in geskikte
formule saam met -3,40 × 103
Final answer/Finale antwoord: 8,88 m
OPTION 1/OPSIE 1
Wnc = K + U Wnc = ½ mvf
2 - ½ mvi2 + mghf - mghi
= ½ m (vf2- vi
2) + mg(hf - hi) -3,40 × 103 = -1 200 + 200(9,8)(hf - 10) h = 8,88 m (8,87765 m)
OPTION 2/OPSIE 2
E(mech/meg)A + Wf = E(mech)B (Ep +Ek)A + Wf = (Ep +Ek) B (mgh + ½mv2) A + Wf = (mgh + ½mv2) B 200(9,8)(10) + ½(200)(42) - 3,40 × 103 = 200(9,8)(h) + ½(200)(2)2
h = 8,88 m (8,87755)
OPTION 3/OPSIE 3
Wnet = K Wf + Ww = ½mvf
2 - ½mvi2
Wf – ΔEp = ½mvf2 - ½mvi
2 Wf - mg(hf - hi) = ½m(vf
2- vi2)
-3,40 × 103 - 200(9,8)(h-10) = -1 200 h = 8,88 m (8,87755 m)
(4)
5.4 OPTION 1 AND 2/OPSIE 1 EN 2: Marking criteria /Nasienriglyne
Appropriate formula/Geskikte formule
Work done by friction/Arbeid verrig deur wrywing
Substitution of/Vervanging van (200)(9,8)(13,12)
Appropriate formula/Geskikte formule
Substitution into power formula/Vervanging in drywingformule
Final answer /Finale antwoord: 1 814,35 W
OPTION 1/OPSIE 1
Wnc = K + U Wengine + Wf = ½mvf
2 - ½ mvi2 + mghf - mghi
= ½m(vf2- vi
2) + mg(hf - hi) Wengine + (50)(15)(2)cos180° = 0 + 200(9,8)(22 – 8,88) Wengine = 27 215,20 J
Pengine = t
Wengine
= 15
20,21527
= 1 814,35 W
Any one/Enige een
Any one/Enige een
Any one/Enige een
Any one/Enige een
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OPTION 2/OPSIE 2
Wnet = K WN + Wengine + Ww + Wf = 0 WN + Wengine - ΔEp + Wf = 0 0 + Wengine - (200)(9,8)(13,12) + (50)(2)(15)cos180° = 0 Wengine = 27 215,20 J OR/OF
Wnet = K WN + Wengine + Ww|| + Wf = 0 WN + Wengine +mgsinθΔxcos180o + Wf = 0
0 + Wengine - (200)(9,8)
x
12,13Δx(-1) + (50)(2)(15)cos180° = 0
Wengine = 27 215,20 J
Pengine = t
Wengine
= 15
20,21525
= 1 814,35 W
OPTION/OPSIE 3: Marking criteria/Nasienriglyne Opsie 3
Appropriate formula/Geskikte formule
Substitution of/Vervanging van - 50
Substitution of/Vervanging van (-200)(9,8)(0,4373) or/of (-200)(9,8)(0,44)
Appropriate formula/Geskikte formule
Substitution into/Vervanging in Pave = Fvave
Final answer/Finale antwoord: 1 814,35 W - 1 824,8 W
OPTION 3/OPSIE 3
Fnet = ma Fengine + Ffriction + Fg// = 0 Fengine + (-50) + (-200)(9,8)(0,4373) = 0 Fengine = 906,52 N (906,52 – 912,4) Pave = Fvave Pave = (908,52)(2) = 1 813,04 W (1 824,8 W) OR/OF W = FengineΔxcosθ = (906,52)(30)cos0o = 27 195,6 J (27 372 W)
P = t
W
=
15
6,19527 = 1 813,04 W (1 824,8 W)
(5) [14]
Any one/Enige een
Any one/Enige een
sinθ =x
h
= )15(2
12,13
= 0,4373
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QUESTION 6/VRAAG 6
6.1 Marking criteria/Nasienriglyne If any of the underlined key words/phrases in the correct context are omitted: - 1 mark per word/phrase. Indien enige van die sleutelwoorde/frases in die korrekte konteks weggelaat word: - 1 punt per woord/frase
The change in frequency (or pitch) (of the sound) detected by a listener because the source and the listener have different velocities relative to the medium of propagation. Die verandering in die frekwensie (of toonhoogte) (van die klank) waargeneem deur 'n luisteraar omdat die bron en die luisteraar verskillende snelhede relatief tot die voortplantingsmedium het. OR/OF An (apparent) change in (observed/detected) frequency (pitch), as a result of the relative motion between a source and an observer (listener). 'n (Skynbare) verandering in (waargenome) frekwensie (toonhoogte), as gevolg van die relatiewe beweging tussen die bron en 'n waarnemer/ luisteraar.
(2) 6.2 Towards/Nader (1)
6.3 s
s
LL f
vv
vvf
OR/OF s
s
L fvv
vf
s
s
fv340
03403148
OR/OFs
s
L fvv
vf
s
s
fv340
03402073
0340
)v340(3148 s
=
0340
)v340(2073 s
vs = 70 m·s-1 (69,95 - 70,16 m·s-1)
(6)
6.4 POSITIVE MARKING FROM QUESTION 6.3 POSITIEWE NASIEN VANAF VRAAG 6.3
OPTION 1/OPSIE 1
v
xt
70
350t
t = 5 s
OPTION 2/OPSIE 2 Δx = viΔt + ½ aΔt2
350 = 70Δt + 0 Δt = 5 s
OPTION 3/OPSIE 3
Δt2
vvΔx fi
Δt2
7070350
Δt = 5 s
(2)
[11]
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QUESTION 7/VRAAG 7
7.1 n =
e
Q
= 19-
-6
101,6)(
104)(
= 2,5 x 1013
(3)
7.2
Electrostatic force on B due to A:/Elektrostatiese krag op B a.g.v. A:
FAB = 2
21
r
QkQ
=
2
-6-69
0,2
)10)(310(410 9
= 2,7 N
(3)
7.3 Marking criteria/Nasienriglyne If any of the underlined key words/phrases in the correct context are omitted: - 1 mark per word/phrase. Indien enige van die sleutelwoorde/frases in die korrekte konteks weggelaat word: - 1 punt per woord/frase
Electric field is a region (in space) where (in which) an (electric) charge experiences a (electric) force. Elektriese veld is 'n gebied (in die ruimte) waarin 'n (elektriese) lading 'n (elektriese) krag ondervind.
(2)
Ignore negative signs Ignoreer negatiewe tekens doeleindes
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7.4
Marking criteria/Nasienriglyne
Appropriate formula/Geskikte formule
Correct substitution for A and B/Korrekte vervanging van A en B
Subtraction of electric fields/Aftrek van elektriesevelde
Final answer/Finale antwoord: 2,3 x106 N∙C-1
OPTION 1/OPSIE 1 Electric field at M due to / Elektriese veld by M as gevolg van: -4 x10-6 C
EAM = 2r
Qk
= 2
-69
)3,0(
)10x(4109
= 4,0 x 105 N∙C-1 (to left /links) Electric field at M due to / Elektriese veld by M as gevolg van: +3 x10-6 C,
EBM = 2r
Qk
= 2
-69
)1,0(
) x10(310x9
= 2,7 x106 N∙C-1 (to right /regs) Net electric field at M /Netto elektrieseveld by M Enet = EBM + EAM = 4,0 x105 – 2,7 x106 = 2,3 x106 N∙C-1
(right/regs) OR/OF Net electric field at M /Netto elektrieseveld by M Enet = EBM + EAM = -4,0 x105 + 2,7 x106
= - 2,3 x106 N∙C-1
= 2,3 x106 N∙C-1 (right)
(5)
OPTION 2/OPSIE 2
FAM = 2
21
r
QkQ =
2
69
)3,0(
Q)104)(109( = 4 x 105Q N
FBM = 2
21
r
QkQ =
2
69
)1,0(
Q)103)(109( = 2,7 x 106Q N
Fnet = 2,7 x 106Q + (-4 x 105Q) = 2,3 x 106Q
E = q
F =
Q
Q103,2 6 = 2,3 x 106 N·C-1 (right/regs)
Physical Sciences/P1/Fisiese Wetenskappe/V1 14 DBE/2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
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7.5 Positive/Positief (1)
7.6 POSITIVE MARKING FROM 7.2/POSITIEWE NASIEN VANAF 7.2
Marking criteria/Nasienriglyne
Correct substitution into Pythagoras’s equation/Korrekte vervanging in Pythagoras se vergelyking
Correct substitution into Coulomb’s Law/Korrekte vervanging in Coulomb se wet
Correct answer/Korrekte antwoord
(3)
(Fnet)2 = (FAD)2 + (FAB)2
(7,69)2 = (FAD)2 + (2,7)2 FAD = 7,2 N
FAD = 2
21
r
QkQ
7,2 = 2
-69
)15,0(
Q)10x4)(10(9
QD = 4,5 x 10-6 C OR/OF
FAD = 2
21
r
QQk
= 2
-69
15,0
Q)10x4(10x9
= 1,6 x 106 Q
Fnet =2
AD
2
AB FF
7,69 = 262 )Q10x6,1(7,2
Q = 4,50 x10-6 C
[17]
OR/OF 2
netF = 2
AD
2
AB FF
Physical Sciences/P1/Fisiese Wetenskappe/V1 15 DBE/2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
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QUESTION 8/VRAAG 8
8.1 Marking criteria/Nasienriglyne If any of the underlined key words/phrases in the correct context are omitted: - 1 mark per word/phrase. Indien enige van die sleutelwoorde/frases in die korrekte konteks weggelaat word: - 1 punt per woord/frase
(Maximum) energy provided (work done) by a battery per coulomb/unit charge passing through it. (Maksimum) energie verskaf (arbeid verrig) deur 'n battery per coulomb/ eenheidslading wat daardeur beweeg.
Work done by the battery to move a unit coulomb of charge across the circuit./Arbeid verrig deur die battery om 'n eenheidslading oor die stroombaan te beweeg.
(2) 8.2 Energy (per coulomb of charge) is converted to heat in the battery due to the
internal resistance. Energie (per coulomb lading) word na hitte omskep binne-in die battery a.g.v. interne weerstand.
(2) 8.3.1
R
VI
0,5
1,5I
= 3 A
(3)
8.3.2 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2
21P R
1+
R
1=
R
1
15
1
25
1
R
1
P
Rp = 9,375
Rext = 9,375 + 4 = 13,38
(13,375 )
21
21p
RR
RRR
1525
)15)(25(Rp
Rp = 9,375
Rext = 9,375 + 4 = 13,38
(13,375 )
(4)
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8.3.3 POSITIVE MARKING FROM QUESTIONS 8.3.1 AND 8.3.2. POSITIEWE NASIEN VANAF VRAAG 8.3.1 EN 8.3.2.
(3)
OPTION 1/OPSIE 1 Ɛ = I(R + r) = 3(13,38 + 0,5) = 41,64 V (Range/Gebied: 41,625 – 41,64)
OPTION 2/OPSIE 2 Ɛ = Vext/eks + Vint = (3)(13,38) + 1,5 = 41,64 V (Range/Gebied: 41,625 – 41,64)
8.4 Yes. /Ja
For the same voltage/potential difference,
a larger current will flow through a smaller resistor (R
VI )
Vir dieselfde spanning/ potensiaalverskil
sal 'n groter stroom deur die kleiner weerstand vloei (R
VI ).
OR/OF
R
1αI , V = constant /konstant
I is inversely proportional to R and V is constant. I is omgekeerd eweredig aan R en V is konstant. OR/OF V|| = IR = (3)(9,38) = 28,14 V
IR2 = R
V=
25
14,28 = 1,13 A
IR3 = R
V=
15
14,28 = 1,88 A
OR/OF V is the same / V is dieselfde
I15Ω = 40
25I
I25Ω = 40
15I
(3) 8.5 Remains the same/Bly dieselfde (1) [18]
Physical Sciences/P1/Fisiese Wetenskappe/V1 17 DBE/2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
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QUESTION 9/VRAAG 9 9.1.1 (DC) motor/(GS-)motor (1) 9.1.2 POSITIVE MARKING FROM QUESTION 9.1.1
POSITIEWE NASIEN VANAF VRAAG 9.1.1 Electrical to mechanical /kinetic (energy) (2 or 0) Elektriese na meganiese/kinetiese (energie) (2 of 0)
(2) 9.1.3 Split ring/commutator/Splitring/kommutator (1) 9.1.4 Anticlockwise/antikloksgewys (2) 9.2.1 (The rms voltage/value of AC is) the AC voltage/potential difference which
dissipates the same amount of energy/heat/power as an equivalent DC voltage/potential difference. (2 or 0) (Die wgk-waarde van WS is) die WS-potensiaalverskil/spanning wat dieselfde hoeveelheid energie/hitte/drywing verbruik as 'n ekwivalente GS-spanning/ potensiaalverskil. (2 of 0) ACCEPT/AANVAAR
The rms voltage/value of AC is the DC potential difference which dissipates the same amount of energy/heat/power as AC. Die wgk-waarde van WS is die GS-potensiaalverskil wat dieselfde hoeveelheid energie/hitte/drywing verbruik as die WS.
(2)
9.2.2 Marking criteria/Nasienriglyne
Appropriate formula for Pave/Geskikte formule vir Pave
Substitution to calculate/Vervanging vir berekening van R
Final answer/Finale antwoord: 242 Ω
OPTION 1/OPSIE 1
R
VP
2
rmsave
R
220=200
2
R = 242 Ω
OPTION 2/OPSIE 2 OPTION 3/OPSIE 3
rmsrmsave IVP
200 = Irms (220) Irms = 0,909 A (0,91)
rmsrmsave IVP
200 = Irms (220) Irms = 0,909 A(0,91)
rms
rms
I
VR or/of
I
VR
909,0
220R R = 242 Ω (241,76 Ω)
Pave = Irms
2R 200 = (0,909)2R
R = 242 Ω
(241,52 Ω)
(3)
Physical Sciences/P1/Fisiese Wetenskappe/V1 18 DBE/2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
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9.2.3 Marking criteria for options 1,2 and 3 /Nasienriglyne vir opsies 1,2 en 3
Appropriate formula to calculate P or Irms /Geskikte formule om P of Irms te bereken
Substitution/Vervanging
Formula for P or W containing t/Formule vir P of W wat t bevat
Substitution/Vervanging
Final answer/Finale antwoord: 55 785,12 J
POSITIVE MARKING FROM QUESTION 9.2.2. POSITIEWE NASIEN VANAF VRAAG 9.2.2.
OPTION 1/OPSIE 1 Marking criteria / Nasienriglyne
Appropriate formula for W containing V/Geskikte formule vir W wat V bevat
Substitution/Vervanging
Final answer/Finale antwoord: 55 785,12 J
(5)
R
tVW
2
242
)60x10()(150
2
= 55 785,12 J
OPTION 2/OPSIE 2
R
V=P
2
rms
ave
242
150=
2
Pav = 92,975 W
Δt
WP
(10)(60)
W92,975
W = 55 785,12 J (55785,12 – 55896 J)
OPTION 3/OPSIE 3 OPTION 4/OPSIE 4
rms
rms
I
VR /
I
VR
rmsI
150 242
rmsI = 0,620 A
rmsrmsave VIP
= (0,62)(150) = 92,97 W (93 W)
Δt
WP
(10)(60)
W92,975
W = 55 785,12 J (55785,12 – 55896 J)
rms
rms
I
VR /
I
VR
rmsI
150 242
rmsI = 0,620 A
W = I2R t = (0,62)2(242)(10)(60) = 55 814,88 J (55785,12 – 55896 J) OR/OF W = VIΔt = (150)(0,62)(600) = 55 800 J
OPTION 5/OPSIE 5
R
V=P
2
rms
ave
242
150=
2
= 92,975 W
Pave= Irms2R
92,975 = Irms2(242)
Irms = 0,6198 A
W = I2RΔt
= (0,6198)2(242)(10)(60)
= 55 778,88 J
[16]
Physical Sciences/P1/Fisiese Wetenskappe/V1 19 DBE/November 2020 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
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QUESTION 10/VRAAG 10 10.1 Photoelectric effect/Fotoëlektriese effek (1) 10.2 Work function (of potassium)/Werksfunksie/Arbeidsfunksie (van kalium) (1) 10.3 Potassium/Kalium
It has the lowest work function / threshold frequency / highest threshold wavelength. Dit het die laagste arbeidsfunksie / drumpelfrekwensie / hoogste drumpel golflengte.
(2)
10.4 Marking criteria/Nasienriglyne If any of the underlined key words/phrases in the correct context are omitted: - 1 mark per word/phrase. Indien enige van die sleutelwoorde/frases in die korrekte konteks weggelaat word: - 1 punt per woord/frase
The work function of a metal is the minimum energy that an electron (in the metal) needs to be emitted/ejected from the metal / surface. Die werkfunksie/arbeidsfunskie van 'n metaal is die minimum energie benodig om 'n elektron vanaf ‘n oppervlak / metaal vry te stel.
(2) 10.5.1 Wo= hfo
= (6,63 x 10-34)(1,75 x 1015) = 1,160 x 10-18 J OR/OF E = Wo + Ek(max) hf = Wo + Ek(max) (6,63 x 10-34)(1,75 x 1015) = Wo + 0 Wo = 1,160 x 10-18 J
(3) 10.5.2 POSITIVE MARKING FROM QUESTION 10.5.1.
POSITIEWE NASIEN VANAF VRAAG 10.5.1.
(max)k0 EWE
2
maxo mv2
1hfhf
(6,63 x 10-34)f = 1,160 x 10-18 + 2
1(9,11 x 10-31) (5,60 x 105)2
f = 1,97 x 1015 Hz
(4)
[13]
TOTAL/TOTAAL: 150
Any one/Enige een
Any one / Enigeen
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